Edits to “Problem”, “Solution”, “0. Write down the given data (a purely f”

Adler edited Jun 17, 2026, 3:03 PM

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		@@ -1,11 +1,76 @@
		−	### ~~Statement~~
		+	### Problem

		−	$6.6.3.$ ~~[Insert the problem statement]~~
		+	$6.6.3.$

		+	The saturated vapor pressure of water at $18^{\circ}C$ is $2 \cdot 10^{3} \ Pa$, and its dielectric constant is 1.0078. From these data, find the average dipole moment of a water molecule in an electric field of strength $10^{3}$ V/m. Reference books give the dipole moment of water as $-0.61 \cdot 10^{-29} C \cdot m$. How can the discrepancy in the results be explained?
		+
			### Solution

		−	12
		+	Before looking at the solution, you can review the first solution method (the second one is also possible, for broadening your knowledge) of problem $6.6.2$ and then try to solve this one.

		−	#### Answer
		+	#### 0. Write down the given data (a purely formal step to make the solution clear).

		−	~~[Insert a concise answer or boxed result]~~
		+	$\varepsilon = 1.0078$
		+
		+	$E = 10^{3} \ V/m$
		+
		+	$T = 18^{\circ}C = 291 \ K$
		+
		+	$\varepsilon_{0} = 8.85 \cdot 10^{-12} \ F/m$
		+
		+	$P = 2 \cdot 10^{3} \ Pa$
		+
		+	$p' = -0.61 \cdot 10^{-29} \ C \cdot m$
		+
		+	$K = 1.38 \cdot 10^{-23} \ J/K$
		+
		+	#### 1. Find the answer to the first question.
		+
		+	We will find the answer by working backwards, that is, we write the formula for the dipole moment and start from there.
		+
		+	\[
		+	\boxed{p=\frac{P_{pol}}{n}}
		+	\]
		+
		+	where $P_{pol}$ is the polarization vector, but since our dielectric is isotropic (the same in all directions), the vector concept does not play a role here, and $n$ is the concentration of molecules; we find it from the basic MKT equation.
		+
		+	$P = nKT \Rightarrow n = \frac{P}{KT}$
		+
		+	$P_{pol}$ is found from the polarization vector formula:
		+
		+	$P_{pol} = \varepsilon_{0}(\varepsilon - 1)E$
		+
		+	Now we find the dipole moment:
		+
		+	\[
		+	\boxed{p = \frac{\varepsilon_{0}(\varepsilon - 1)EKT}{P}}
		+	\]
		+
		+	$p = \frac{8.85 \cdot 10^{-12} \cdot (1.0078 - 1) \cdot 10^{3} \cdot 1.38 \cdot 10^{-23} \cdot 291}{2 \cdot 10^{3}} \approx 1.4 \cdot 10^{-34} \ C \cdot m$
		+
		+	#### 2. Reasoning for the answer to the second question.
		+
		+	First of all, we need to understand what we found and what is given in reference materials.
		+
		+	We found the induced dipole moment, while reference books give the permanent dipole moment of the molecule.
		+
		+	What is the difference between the induced and the permanent dipole moment?
		+
		+	The induced dipole moment is a temporary dipole moment that arises in a particle under the influence of an external electric field.
		+
		+	The permanent dipole moment remains with the molecule forever (as the name suggests), even at $E = 0$, in our equations.
		+
		+	If we speak quantitatively about the difference, the Langevin-Debye formula helps, which has the form:
		+
		+	\[
		+	\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n}{3\varepsilon_{0}}\left(\alpha + \frac{p_{0}^{2}}{3KT}\right)}
		+	\]
		+
		+	Here $p_{0}^{2}$ is precisely that permanent dipole moment, and $\alpha$ is the electronic polarizability.
		+
		+	More detailed information about this formula can be read here: https://www.sciencedirect.com/topics/pharmacology-toxicology-and-pharmaceutical-science/lachesine
		+	### Answer
		+
		+	\[
		+	\boxed{p = 1.4 \cdot 10^{-34} \ C \cdot m}
		+	\]

	@@ -1,11 +1,76 @@
	### ~~Statement~~		### Problem

	$6.6.3.$ ~~[Insert the problem statement]~~		$6.6.3.$

			The saturated vapor pressure of water at $18^{\circ}C$ is $2 \cdot 10^{3} \ Pa$, and its dielectric constant is 1.0078. From these data, find the average dipole moment of a water molecule in an electric field of strength $10^{3}$ V/m. Reference books give the dipole moment of water as $-0.61 \cdot 10^{-29} C \cdot m$. How can the discrepancy in the results be explained?

	### Solution		### Solution

	12		Before looking at the solution, you can review the first solution method (the second one is also possible, for broadening your knowledge) of problem $6.6.2$ and then try to solve this one.

	#### Answer		#### 0. Write down the given data (a purely formal step to make the solution clear).

	~~[Insert a concise answer or boxed result]~~		$\varepsilon = 1.0078$

			$E = 10^{3} \ V/m$

			$T = 18^{\circ}C = 291 \ K$

			$\varepsilon_{0} = 8.85 \cdot 10^{-12} \ F/m$

			$P = 2 \cdot 10^{3} \ Pa$

			$p' = -0.61 \cdot 10^{-29} \ C \cdot m$

			$K = 1.38 \cdot 10^{-23} \ J/K$

			#### 1. Find the answer to the first question.

			We will find the answer by working backwards, that is, we write the formula for the dipole moment and start from there.

			\[
			\boxed{p=\frac{P_{pol}}{n}}
			\]

			where $P_{pol}$ is the polarization vector, but since our dielectric is isotropic (the same in all directions), the vector concept does not play a role here, and $n$ is the concentration of molecules; we find it from the basic MKT equation.

			$P = nKT \Rightarrow n = \frac{P}{KT}$

			$P_{pol}$ is found from the polarization vector formula:

			$P_{pol} = \varepsilon_{0}(\varepsilon - 1)E$

			Now we find the dipole moment:

			\[
			\boxed{p = \frac{\varepsilon_{0}(\varepsilon - 1)EKT}{P}}
			\]

			$p = \frac{8.85 \cdot 10^{-12} \cdot (1.0078 - 1) \cdot 10^{3} \cdot 1.38 \cdot 10^{-23} \cdot 291}{2 \cdot 10^{3}} \approx 1.4 \cdot 10^{-34} \ C \cdot m$

			#### 2. Reasoning for the answer to the second question.

			First of all, we need to understand what we found and what is given in reference materials.

			We found the induced dipole moment, while reference books give the permanent dipole moment of the molecule.

			What is the difference between the induced and the permanent dipole moment?

			The induced dipole moment is a temporary dipole moment that arises in a particle under the influence of an external electric field.

			The permanent dipole moment remains with the molecule forever (as the name suggests), even at $E = 0$, in our equations.

			If we speak quantitatively about the difference, the Langevin-Debye formula helps, which has the form:

			\[
			\boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n}{3\varepsilon_{0}}\left(\alpha + \frac{p_{0}^{2}}{3KT}\right)}
			\]

			Here $p_{0}^{2}$ is precisely that permanent dipole moment, and $\alpha$ is the electronic polarizability.

			More detailed information about this formula can be read here: https://www.sciencedirect.com/topics/pharmacology-toxicology-and-pharmaceutical-science/lachesine
			### Answer

			\[
			\boxed{p = 1.4 \cdot 10^{-34} \ C \cdot m}
			\]